Since the ball starts from rest, we know thatT1=0. Also, setting the origin of our coordinate system atthe ground, we have0+mg(H+h)=12mV2+mghSolving forVgivesV=02gHSince the initial velocity of the ball isv=−Vk, there followsvt=t·v=Vsinβvn=n·v=−VcosβTurning to the impact between ball and the incline, the tangential velocity component is unchanged, whichtells us thatvIt=VsinβwherevI=vInn+vIttis the ball’s velocity just after the impact. The impact relation tells us that since thecoefficient of restitution ise, and since the incline is immovable,0−vIn=e(vn−0)=⇒vIn=−evn=eVcosβHence, just after the impact, the ball’s velocity isvI=eVcosβn+VsinβtSubstituting fornandtfrom above, we havevI=eVcosβ(isinβ+kcosβ)+Vsinβ(icosβ−ksinβ)=(eVsinβcosβ+Vsinβcosβ)i+DeVcos2β−Vsin2βikMaking use of the following trigonometric identitiessinβcosβ
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This note was uploaded on 05/12/2010 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.